Question: Find $\lim_{x\to \frac\pi2}\dfrac{-\cos(x)}{\left(x-\dfrac\pi2\right)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-2\pi$ (Choice B) B $-1$ (Choice C) C $1$ (Choice D) D The limit doesn't exist.
Explanation: Substituting $x=\dfrac\pi 2$ into $\dfrac{-\cos(x)}{\left(x-\dfrac\pi2\right)}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to \frac\pi2}\dfrac{-\cos(x)}{\left(x-\dfrac\pi2\right)} \\\\ &=\lim_{x\to \frac\pi2}\dfrac{\dfrac{d}{dx}[-\cos(x)]}{\dfrac{d}{dx}\left[x-\dfrac\pi2\right]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to \frac\pi2}\dfrac{\sin(x)}{1} \\\\ &=\dfrac{\sin\left(\dfrac\pi2\right)}{1} \gray{\text{Substitution}} \\\\ &=1 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to \frac\pi2}\dfrac{\dfrac{d}{dx}[-\cos(x)]}{\dfrac{d}{dx}\left[x-\dfrac\pi2\right]}$ actually exists. In conclusion, $\lim_{x\to \frac\pi2}\dfrac{-\cos(x)}{\left(x-\dfrac\pi2\right)}=1$.